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Diverging Integral with Bessel Function

How to compute this integral of Bessel functions?Compute a Real Integral using Residue TheoremInverting a Characteristic Function for half-cubic Student's T entailing a Modified Bessel of 2nd kindDefinite integral of Bessel function product over fourth powerEvaluating functions similar to the Bessel functionsContour integral to real integral: find suitable change of variablesFinishing off integral of a Bessel functionIntegral with an inverse function limitDifferentiating an improper integral with a variable limitDoubt In A simple integral using complex numbers

0

$begingroup$

I am looking for the solution to the integral:

$$int_a^infty x J_n(alpha x);dx$$

where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.

---

The original integral of interest is actually this one

$$int_0^infty x J_n(alpha x);dx$$

However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...

I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.

integration definite-integrals bessel-functions parameter-estimation

share|cite|improve this question

edited Mar 10 at 20:10

Robert Howard

2,2112933

asked Jan 25 at 23:16

MsTaisMsTais

1808

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^inftyxJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
  $endgroup$
  – jmerry
  Jan 25 at 23:47
  

  
  
  
  

add a comment | 

0

$begingroup$

I am looking for the solution to the integral:

$$int_a^infty x J_n(alpha x);dx$$

where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.

---

The original integral of interest is actually this one

$$int_0^infty x J_n(alpha x);dx$$

However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...

I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.

integration definite-integrals bessel-functions parameter-estimation

share|cite|improve this question

edited Mar 10 at 20:10

Robert Howard

2,2112933

asked Jan 25 at 23:16

MsTaisMsTais

1808

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^inftyxJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
  $endgroup$
  – jmerry
  Jan 25 at 23:47
  

  
  
  
  

add a comment | 

$begingroup$

I am looking for the solution to the integral:

$$int_a^infty x J_n(alpha x);dx$$

where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.

---

The original integral of interest is actually this one

$$int_0^infty x J_n(alpha x);dx$$

However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...

I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.

integration definite-integrals bessel-functions parameter-estimation

share|cite|improve this question

edited Mar 10 at 20:10

Robert Howard

2,2112933

asked Jan 25 at 23:16

MsTaisMsTais

1808

$endgroup$

share|cite|improve this question

edited Mar 10 at 20:10

Robert Howard

2,2112933

asked Jan 25 at 23:16

MsTaisMsTais

1808

share|cite|improve this question

edited Mar 10 at 20:10

Robert Howard

2,2112933

asked Jan 25 at 23:16

MsTaisMsTais

1808

edited Mar 10 at 20:10

Robert Howard

2,2112933

edited Mar 10 at 20:10

Robert Howard

2,2112933

asked Jan 25 at 23:16

MsTaisMsTais

1808

asked Jan 25 at 23:16

MsTaisMsTais

1808

1 Answer
1

active

oldest

votes

2

$begingroup$

Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.

The Hankel Transform of order n, and its inverse, can be defined as

$$mathscrH_nleftf(r)right = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscrH^-1_nleftF(q)right = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$

Note that the Hankel Transform is its own inverse.

For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:

$$mathscrH^-1_0leftdfracdelta(q)qright = int_0^infty dfracdelta(q)qJ_0(qr)qspace dq = 1$$
$$mathscrH_0left1right = int_0^infty J_0(qr)rspace dr = dfracdelta(q)q$$

For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.

share|cite|improve this answer

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139

$endgroup$

add a comment | 

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2

$begingroup$

Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.

The Hankel Transform of order n, and its inverse, can be defined as

$$mathscrH_nleftf(r)right = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscrH^-1_nleftF(q)right = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$

Note that the Hankel Transform is its own inverse.

For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:

$$mathscrH^-1_0leftdfracdelta(q)qright = int_0^infty dfracdelta(q)qJ_0(qr)qspace dq = 1$$
$$mathscrH_0left1right = int_0^infty J_0(qr)rspace dr = dfracdelta(q)q$$

For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.

share|cite|improve this answer

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139

$endgroup$

add a comment | 

2

$begingroup$

Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.

The Hankel Transform of order n, and its inverse, can be defined as

$$mathscrH_nleftf(r)right = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscrH^-1_nleftF(q)right = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$

Note that the Hankel Transform is its own inverse.

For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:

$$mathscrH^-1_0leftdfracdelta(q)qright = int_0^infty dfracdelta(q)qJ_0(qr)qspace dq = 1$$
$$mathscrH_0left1right = int_0^infty J_0(qr)rspace dr = dfracdelta(q)q$$

For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.

share|cite|improve this answer

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139

$endgroup$

add a comment | 

$begingroup$

Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.

The Hankel Transform of order n, and its inverse, can be defined as

$$mathscrH_nleftf(r)right = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscrH^-1_nleftF(q)right = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$

Note that the Hankel Transform is its own inverse.

For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:

$$mathscrH^-1_0leftdfracdelta(q)qright = int_0^infty dfracdelta(q)qJ_0(qr)qspace dq = 1$$
$$mathscrH_0left1right = int_0^infty J_0(qr)rspace dr = dfracdelta(q)q$$

For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.

share|cite|improve this answer

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139

$endgroup$

share|cite|improve this answer

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139

answered Jan 26 at 0:09

Andy WallsAndy Walls

1,754139